How to remove all duplicates from an array of objects?

Question

I have an object that contains an array of objects.

obj = {};

obj.arr = new Array();

obj.arr.push({place:"here",name:"stuff"});
obj.arr.push({place:"there",name:"morestuff"});
obj.arr.push({place:"there",name:"morestuff"});

I'm wondering what is the best method to remove duplicate objects from an array. So for example, obj.arr would become...

{place:"here",name:"stuff"},
{place:"there",name:"morestuff"}

Answer 1

How about with some ES6 magic?

obj.arr = obj.arr.filter((value, index, self) =>
  index === self.findIndex((t) => (
    t.place === value.place && t.name === value.name
  ))
)

Reference URL

A more generic solution would be:

const uniqueArray = obj.arr.filter((value, index) => {
  const _value = JSON.stringify(value);
  return index === obj.arr.findIndex(obj => {
    return JSON.stringify(obj) === _value;
  });
});

Using the above property strategy instead of JSON.stringify:

const isPropValuesEqual = (subject, target, propNames) =>
  propNames.every(propName => subject[propName] === target[propName]);

const getUniqueItemsByProperties = (items, propNames) => 
  items.filter((item, index, array) =>
    index === array.findIndex(foundItem => isPropValuesEqual(foundItem, item, propNames))
  );

You can add a wrapper if you want the propNames property to be either an array or a value:

const getUniqueItemsByProperties = (items, propNames) => {
  const propNamesArray = Array.from(propNames);

  return items.filter((item, index, array) =>
    index === array.findIndex(foundItem => isPropValuesEqual(foundItem, item, propNamesArray))
  );
};

allowing both getUniqueItemsByProperties('a') and getUniqueItemsByProperties(['a']);

Stackblitz Example

Explanation

• Start by understanding the two methods used:
  • filter, findIndex
• Next take your idea of what makes your two objects equal and keep that in mind.
• We can detect something as a duplicate, if it satisfies the criterion that we have just thought of, but its position is not at the first instance of an object with the criterion.
• Therefore we can use the above criterion to determine if something is a duplicate.

Answer 2

One liners with filter ( Preserves order )

Find unique id's in an array.

arr.filter((v,i,a)=>a.findIndex(v2=>(v2.id===v.id))===i)

If the order is not important, map solutions will be faster: Solution with map

---

Unique by multiple properties ( place and name )

arr.filter((v,i,a)=>a.findIndex(v2=>['place','name'].every(k=>v2[k] ===v[k]))===i)

Unique by all properties (This will be slow for large arrays)

arr.filter((v,i,a)=>a.findIndex(v2=>(JSON.stringify(v2) === JSON.stringify(v)))===i)

Keep the last occurrence by replacing findIndex with findLastIndex.

arr.filter((v,i,a)=>a.findLastIndex(v2=>(v2.place === v.place))===i)

Answer 3

Using ES6+ in a single line you can get a unique list of objects by key:

const key = 'place';
const unique = [...new Map(arr.map(item => [item[key], item])).values()]

It can be put into a function:

function getUniqueListBy(arr, key) {
    return [...new Map(arr.map(item => [item[key], item])).values()]
}

Here is a working example:

const arr = [
    {place: "here",  name: "x", other: "other stuff1" },
    {place: "there", name: "x", other: "other stuff2" },
    {place: "here",  name: "y", other: "other stuff4" },
    {place: "here",  name: "z", other: "other stuff5" }
]

function getUniqueListBy(arr, key) {
    return [...new Map(arr.map(item => [item[key], item])).values()]
}

const arr1 = getUniqueListBy(arr, 'place')

console.log("Unique by place")
console.log(JSON.stringify(arr1))

console.log("\nUnique by name")
const arr2 = getUniqueListBy(arr, 'name')

console.log(JSON.stringify(arr2))

How does it work

First the array is remapped in a way that it can be used as an input for a Map.

arr.map(item => [item[key], item]);

which means each item of the array will be transformed in another array with 2 elements; the selected key as first element and the entire initial item as second element, this is called an entry (ex. array entries, map entries). And here is the official doc with an example showing how to add array entries in Map constructor.

Example when key is place:

[["here", {place: "here",  name: "x", other: "other stuff1" }], ...]

Secondly, we pass this modified array to the Map constructor and here is the magic happening. Map will eliminate the duplicate keys values, keeping only last inserted value of the same key. Note: Map keeps the order of insertion. (check difference between Map and object)

new Map(entry array just mapped above)

Third we use the map values to retrieve the original items, but this time without duplicates.

new Map(mappedArr).values()

And last one is to add those values into a fresh new array so that it can look as the initial structure and return that:

return [...new Map(mappedArr).values()]

Answer 4

Simple and performant solution with a better runtime than the 70+ answers that already exist:

const ids = arr.map(({ id }) => id);
const filtered = arr.filter(({ id }, index) => !ids.includes(id, index + 1));

---

Example:

const arr = [{
  id: 1,
  name: 'one'
}, {
  id: 2,
  name: 'two'
}, {
  id: 1,
  name: 'one'
}];

const ids = arr.map(({ id }) => id);
const filtered = arr.filter(({ id }, index) => !ids.includes(id, index + 1));

console.log(filtered);

How it works:

Array.filter() removes all duplicate objects by checking if the previously mapped id-array includes the current id ({id} destructs the object into only its id). To only filter out actual duplicates, it is using Array.includes()'s second parameter fromIndex with index + 1 which will ignore the current object and all previous.

Since every iteration of the filter callback method will only search the array beginning at the current index + 1, this also dramatically reduces the runtime because only objects not previously filtered get checked.

What if you don't have a single unique identifier like id?

Just create a temporary one:

const objToId = ({ name, city, birthyear }) => `${name}-${city}-${birthyear}`;


const ids = arr.map(objToId);
const filtered = arr.filter((item, index) => !ids.includes(objToId(item), index + 1));

Answer 5

A primitive method would be:

const obj = {};

for (let i = 0, len = things.thing.length; i < len; i++) {
  obj[things.thing[i]['place']] = things.thing[i];
}

things.thing = new Array();

 for (const key in obj) { 
   things.thing.push(obj[key]);
}

Answer 6

If you can use Javascript libraries such as underscore or lodash, I recommend having a look at _.uniq function in their libraries. From lodash:

_.uniq(array, [isSorted=false], [callback=_.identity], [thisArg])

Basically, you pass in the array that in here is an object literal and you pass in the attribute that you want to remove duplicates with in the original data array, like this:

var data = [{'name': 'Amir', 'surname': 'Rahnama'}, {'name': 'Amir', 'surname': 'Stevens'}];
var non_duplidated_data = _.uniq(data, 'name'); 

UPDATE: Lodash now has introduced a .uniqBy as well.

Answer 7

I had this exact same requirement, to remove duplicate objects in a array, based on duplicates on a single field. I found the code here: Javascript: Remove Duplicates from Array of Objects

So in my example, I'm removing any object from the array that has a duplicate licenseNum string value.

var arrayWithDuplicates = [
    {"type":"LICENSE", "licenseNum": "12345", state:"NV"},
    {"type":"LICENSE", "licenseNum": "A7846", state:"CA"},
    {"type":"LICENSE", "licenseNum": "12345", state:"OR"},
    {"type":"LICENSE", "licenseNum": "10849", state:"CA"},
    {"type":"LICENSE", "licenseNum": "B7037", state:"WA"},
    {"type":"LICENSE", "licenseNum": "12345", state:"NM"}
];

function removeDuplicates(originalArray, prop) {
     var newArray = [];
     var lookupObject  = {};

     for(var i in originalArray) {
        lookupObject[originalArray[i][prop]] = originalArray[i];
     }

     for(i in lookupObject) {
         newArray.push(lookupObject[i]);
     }
      return newArray;
 }

var uniqueArray = removeDuplicates(arrayWithDuplicates, "licenseNum");
console.log("uniqueArray is: " + JSON.stringify(uniqueArray));

The results:

uniqueArray is:

[{"type":"LICENSE","licenseNum":"10849","state":"CA"},
{"type":"LICENSE","licenseNum":"12345","state":"NM"},
{"type":"LICENSE","licenseNum":"A7846","state":"CA"},
{"type":"LICENSE","licenseNum":"B7037","state":"WA"}]

Answer 8

One liner using Set

var things = new Object();

things.thing = new Array();

things.thing.push({place:"here",name:"stuff"});
things.thing.push({place:"there",name:"morestuff"});
things.thing.push({place:"there",name:"morestuff"});

// assign things.thing to myData for brevity
var myData = things.thing;

things.thing = Array.from(new Set(myData.map(JSON.stringify))).map(JSON.parse);

console.log(things.thing)

Explanation:

1. new Set(myData.map(JSON.stringify)) creates a Set object using the stringified myData elements.
2. Set object will ensure that every element is unique.
3. Then I create an array based on the elements of the created set using Array.from.
4. Finally, I use JSON.parse to convert stringified element back to an object.

Answer 9

ES6 one liner is here

let arr = [
  {id:1,name:"sravan ganji"},
  {id:2,name:"pinky"},
  {id:4,name:"mammu"},
  {id:3,name:"avy"},
  {id:3,name:"rashni"},
];

console.log(Object.values(arr.reduce((acc,cur)=>Object.assign(acc,{[cur.id]:cur}),{})))

Answer 10

To remove all duplicates from an array of objects, the simplest way is use filter:

var uniq = {};
var arr  = [{"id":"1"},{"id":"1"},{"id":"2"}];
var arrFiltered = arr.filter(obj => !uniq[obj.id] && (uniq[obj.id] = true));
console.log('arrFiltered', arrFiltered);

Answer 11

One liners with Map ( High performance, Does not preserve order )

Find unique id's in array arr.

const arrUniq = [...new Map(arr.map(v => [v.id, v])).values()]

If the order is important check out the solution with filter: Solution with filter

---

Unique by multiple properties ( place and name ) in array arr

const arrUniq = [...new Map(arr.map(v => [JSON.stringify([v.place,v.name]), v])).values()]

Unique by all properties in array arr

const arrUniq = [...new Map(arr.map(v => [JSON.stringify(v), v])).values()]

Keep the first occurrence in array arr

const arrUniq = [...new Map(arr.slice().reverse().map(v => [v.id, v])).values()].reverse()

Answer 12

Here's another option to do it using Array iterating methods if you need comparison only by one field of an object:

    function uniq(a, param){
        return a.filter(function(item, pos, array){
            return array.map(function(mapItem){ return mapItem[param]; }).indexOf(item[param]) === pos;
        })
    }

    uniq(things.thing, 'place');

Answer 13

This is a generic way of doing this: you pass in a function that tests whether two elements of an array are considered equal. In this case, it compares the values of the name and place properties of the two objects being compared.

ES5 answer

function removeDuplicates(arr, equals) {
    var originalArr = arr.slice(0);
    var i, len, val;
    arr.length = 0;

    for (i = 0, len = originalArr.length; i < len; ++i) {
        val = originalArr[i];
        if (!arr.some(function(item) { return equals(item, val); })) {
            arr.push(val);
        }
    }
}

function thingsEqual(thing1, thing2) {
    return thing1.place === thing2.place
        && thing1.name === thing2.name;
}

var things = [
  {place:"here",name:"stuff"},
  {place:"there",name:"morestuff"},
  {place:"there",name:"morestuff"}
];

removeDuplicates(things, thingsEqual);
console.log(things);

Original ES3 answer

function arrayContains(arr, val, equals) {
    var i = arr.length;
    while (i--) {
        if ( equals(arr[i], val) ) {
            return true;
        }
    }
    return false;
}

function removeDuplicates(arr, equals) {
    var originalArr = arr.slice(0);
    var i, len, j, val;
    arr.length = 0;

    for (i = 0, len = originalArr.length; i < len; ++i) {
        val = originalArr[i];
        if (!arrayContains(arr, val, equals)) {
            arr.push(val);
        }
    }
}

function thingsEqual(thing1, thing2) {
    return thing1.place === thing2.place
        && thing1.name === thing2.name;
}

removeDuplicates(things.thing, thingsEqual);

Answer 14

If you can wait to eliminate the duplicates until after all the additions, the typical approach is to first sort the array and then eliminate duplicates. The sorting avoids the N * N approach of scanning the array for each element as you walk through them.

The "eliminate duplicates" function is usually called unique or uniq. Some existing implementations may combine the two steps, e.g., prototype's uniq

This post has few ideas to try (and some to avoid :-) ) if your library doesn't already have one! Personally I find this one the most straight forward:

    function unique(a){
        a.sort();
        for(var i = 1; i < a.length; ){
            if(a[i-1] == a[i]){
                a.splice(i, 1);
            } else {
                i++;
            }
        }
        return a;
    }  

    // Provide your own comparison
    function unique(a, compareFunc){
        a.sort( compareFunc );
        for(var i = 1; i < a.length; ){
            if( compareFunc(a[i-1], a[i]) === 0){
                a.splice(i, 1);
            } else {
                i++;
            }
        }
        return a;
    }

Answer 15

I think the best approach is using reduce and Map object. This is a single line solution.

const data = [
  {id: 1, name: 'David'},
  {id: 2, name: 'Mark'},
  {id: 2, name: 'Lora'},
  {id: 4, name: 'Tyler'},
  {id: 4, name: 'Donald'},
  {id: 5, name: 'Adrian'},
  {id: 6, name: 'Michael'}
]

const uniqueData = [...data.reduce((map, obj) => map.set(obj.id, obj), new Map()).values()];

console.log(uniqueData)

/*
  in `map.set(obj.id, obj)`
  
  'obj.id' is key. (don't worry. we'll get only values using the .values() method)
  'obj' is whole object.
*/

Answer 16

Considering lodash.uniqWith

const objects = [{ 'x': 1, 'y': 2 }, { 'x': 2, 'y': 1 }, { 'x': 1, 'y': 2 }];
 
_.uniqWith(objects, _.isEqual);
// => [{ 'x': 1, 'y': 2 }, { 'x': 2, 'y': 1 }]

Answer 17

To add one more to the list. Using ES6 and Array.reduce with Array.find.
In this example filtering objects based on a guid property.

let filtered = array.reduce((accumulator, current) => {
  if (! accumulator.find(({guid}) => guid === current.guid)) {
    accumulator.push(current);
  }
  return accumulator;
}, []);

Extending this one to allow selection of a property and compress it into a one liner:

const uniqify = (array, key) => array.reduce((prev, curr) => prev.find(a => a[key] === curr[key]) ? prev : prev.push(curr) && prev, []);

To use it pass an array of objects and the name of the key you wish to de-dupe on as a string value:

const result = uniqify(myArrayOfObjects, 'guid')

Answer 18

You could also use a Map:

const dedupThings = Array.from(things.thing.reduce((m, t) => m.set(t.place, t), new Map()).values());

Full sample:

const things = new Object();

things.thing = new Array();

things.thing.push({place:"here",name:"stuff"});
things.thing.push({place:"there",name:"morestuff"});
things.thing.push({place:"there",name:"morestuff"});

const dedupThings = Array.from(things.thing.reduce((m, t) => m.set(t.place, t), new Map()).values());

console.log(JSON.stringify(dedupThings, null, 4));

Result:

[
    {
        "place": "here",
        "name": "stuff"
    },
    {
        "place": "there",
        "name": "morestuff"
    }
]

Answer 19

Dang, kids, let's crush this thing down, why don't we?

let uniqIds = {}, source = [{id:'a'},{id:'b'},{id:'c'},{id:'b'},{id:'a'},{id:'d'}];
let filtered = source.filter(obj => !uniqIds[obj.id] && (uniqIds[obj.id] = true));
console.log(filtered);
// EXPECTED: [{id:'a'},{id:'b'},{id:'c'},{id:'d'}];

Answer 20

let myData = [{place:"here",name:"stuff"}, 
 {place:"there",name:"morestuff"},
 {place:"there",name:"morestuff"}];


let q = [...new Map(myData.map(obj => [JSON.stringify(obj), obj])).values()];

console.log(q)

One-liner using ES6 and new Map().

// assign things.thing to myData
let myData = things.thing;

[...new Map(myData.map(obj => [JSON.stringify(obj), obj])).values()];

Details:-

1. Doing .map() on the data list and converting each individual object into a [key, value] pair array(length =2), the first element(key) would be the stringified version of the object and second(value) would be an object itself.
2. Adding above created array list to new Map() would have the key as stringified object and any same key addition would result in overriding the already existing key.
3. Using .values() would give MapIterator with all values in a Map (obj in our case)
4. Finally, spread ... operator to give new Array with values from the above step.

Answer 21

A TypeScript solution

This will remove duplicate objects and also preserve the types of the objects.

function removeDuplicateObjects(array: any[]) {
  return [...new Set(array.map(s => JSON.stringify(s)))]
    .map(s => JSON.parse(s));
}

Answer 22

 const things = [
  {place:"here",name:"stuff"},
  {place:"there",name:"morestuff"},
  {place:"there",name:"morestuff"}
];
const filteredArr = things.reduce((thing, current) => {
  const x = thing.find(item => item.place === current.place);
  if (!x) {
    return thing.concat([current]);
  } else {
    return thing;
  }
}, []);
console.log(filteredArr)

Solution Via Set Object | According to the data type

const seen = new Set();
 const things = [
  {place:"here",name:"stuff"},
  {place:"there",name:"morestuff"},
  {place:"there",name:"morestuff"}
];

const filteredArr = things.filter(el => {
  const duplicate = seen.has(el.place);
  seen.add(el.place);
  return !duplicate;
});
console.log(filteredArr)

Set Object Feature

Each value in the Set Object has to be unique, the value equality will be checked

The Purpose of Set object storing unique values according to the Data type , whether primitive values or object references.it has very useful four Instance methods add, clear , has & delete.

Unique & data Type feature:..

addmethod

it's push unique data into collection by default also preserve data type .. that means it prevent to push duplicate item into collection also it will check data type by default...

has method

sometime needs to check data item exist into the collection and . it's handy method for the collection to cheek unique id or item and data type..

delete method

it will remove specific item from the collection by identifying data type..

clear method

it will remove all collection items from one specific variable and set as empty object

Set object has also Iteration methods & more feature..

Better Read from Here : Set - JavaScript | MDN

Answer 23

removeDuplicates() takes in an array of objects and returns a new array without any duplicate objects (based on the id property).

const allTests = [
  {name: 'Test1', id: '1'}, 
  {name: 'Test3', id: '3'},
  {name: 'Test2', id: '2'},
  {name: 'Test2', id: '2'},
  {name: 'Test3', id: '3'}
];

function removeDuplicates(array) {
  let uniq = {};
  return array.filter(obj => !uniq[obj.id] && (uniq[obj.id] = true))
}

removeDuplicates(allTests);

Expected outcome:

[
  {name: 'Test1', id: '1'}, 
  {name: 'Test3', id: '3'},
  {name: 'Test2', id: '2'}
];

First, we set the value of variable uniq to an empty object.

Next, we filter through the array of objects. Filter creates a new array with all elements that pass the test implemented by the provided function.

return array.filter(obj => !uniq[obj.id] && (uniq[obj.id] = true));

Above, we use the short-circuiting functionality of &&. If the left side of the && evaluates to true, then it returns the value on the right of the &&. If the left side is false, it returns what is on the left side of the &&.

For each object(obj) we check uniq for a property named the value of obj.id (In this case, on the first iteration it would check for the property '1'.) We want the opposite of what it returns (either true or false) which is why we use the ! in !uniq[obj.id]. If uniq has the id property already, it returns true which evaluates to false (!) telling the filter function NOT to add that obj. However, if it does not find the obj.id property, it returns false which then evaluates to true (!) and returns everything to the right of the &&, or (uniq[obj.id] = true). This is a truthy value, telling the filter method to add that obj to the returned array, and it also adds the property {1: true} to uniq. This ensures that any other obj instance with that same id will not be added again.

Answer 24

Fast (less runtime) and type-safe answer for lazy Typescript developers:

export const uniqueBy = <T>( uniqueKey: keyof T, objects: T[]): T[] => {
  const ids = objects.map(object => object[uniqueKey]);
  return objects.filter((object, index) => !ids.includes(object[uniqueKey], index + 1));
} 

Answer 25

This way works well for me:

function arrayUnique(arr, uniqueKey) {
  const flagList = new Set()
  return arr.filter(function(item) {
    if (!flagList.has(item[uniqueKey])) {
      flagList.add(item[uniqueKey])
      return true
    }
  })
}
const data = [
  {
    name: 'Kyle',
    occupation: 'Fashion Designer'
  },
  {
    name: 'Kyle',
    occupation: 'Fashion Designer'
  },
  {
    name: 'Emily',
    occupation: 'Web Designer'
  },
  {
    name: 'Melissa',
    occupation: 'Fashion Designer'
  },
  {
    name: 'Tom',
    occupation: 'Web Developer'
  },
  {
    name: 'Tom',
    occupation: 'Web Developer'
  }
]
console.table(arrayUnique(data, 'name'))// work well

printout

┌─────────┬───────────┬────────────────────┐
│ (index) │   name    │     occupation     │
├─────────┼───────────┼────────────────────┤
│    0    │  'Kyle'   │ 'Fashion Designer' │
│    1    │  'Emily'  │   'Web Designer'   │
│    2    │ 'Melissa' │ 'Fashion Designer' │
│    3    │   'Tom'   │  'Web Developer'   │
└─────────┴───────────┴────────────────────┘

ES5:

function arrayUnique(arr, uniqueKey) {
  const flagList = []
  return arr.filter(function(item) {
    if (flagList.indexOf(item[uniqueKey]) === -1) {
      flagList.push(item[uniqueKey])
      return true
    }
  })
}

These two ways are simpler and more understandable.

Answer 26

Here is a solution for ES6 where you only want to keep the last item. This solution is functional and Airbnb style compliant.

const things = {
  thing: [
    { place: 'here', name: 'stuff' },
    { place: 'there', name: 'morestuff1' },
    { place: 'there', name: 'morestuff2' }, 
  ],
};

const removeDuplicates = (array, key) => {
  return array.reduce((arr, item) => {
    const removed = arr.filter(i => i[key] !== item[key]);
    return [...removed, item];
  }, []);
};

console.log(removeDuplicates(things.thing, 'place'));
// > [{ place: 'here', name: 'stuff' }, { place: 'there', name: 'morestuff2' }]

Answer 27

If array contains objects, then you can use this to remove duplicate

const persons= [
      { id: 1, name: 'John',phone:'23' },
      { id: 2, name: 'Jane',phone:'23'},
      { id: 1, name: 'Johnny',phone:'56' },
      { id: 4, name: 'Alice',phone:'67' },
    ];
const unique = [...new Map(persons.map((m) => [m.id, m])).values()];

if remove duplicates on the basis of phone, just replace m.id with m.phone

const unique = [...new Map(persons.map((m) => [m.phone, m])).values()];

Answer 28

I know there is a ton of answers in this question already, but bear with me...

Some of the objects in your array may have additional properties that you are not interested in, or you simply want to find the unique objects considering only a subset of the properties.

Consider the array below. Say you want to find the unique objects in this array considering only propOne and propTwo, and ignore any other properties that may be there.

The expected result should include only the first and last objects. So here goes the code:

const array = [{
    propOne: 'a',
    propTwo: 'b',
    propThree: 'I have no part in this...'
},
{
    propOne: 'a',
    propTwo: 'b',
    someOtherProperty: 'no one cares about this...'
},
{
    propOne: 'x',
    propTwo: 'y',
    yetAnotherJunk: 'I am valueless really',
    noOneHasThis: 'I have something no one has'
}];

const uniques = [...new Set(
    array.map(x => JSON.stringify(((o) => ({
        propOne: o.propOne,
        propTwo: o.propTwo
    }))(x))))
].map(JSON.parse);

console.log(uniques);

Answer 29

Another option would be to create a custom indexOf function, which compares the values of your chosen property for each object and wrap this in a reduce function.

var uniq = redundant_array.reduce(function(a,b){
      function indexOfProperty (a, b){
          for (var i=0;i<a.length;i++){
              if(a[i].property == b.property){
                   return i;
               }
          }
         return -1;
      }

      if (indexOfProperty(a,b) < 0 ) a.push(b);
        return a;
    },[]);

Answer 30

This solution worked best for me , by utilising Array.from Method, And also its shorter and readable.

let person = [
{name: "john"}, 
{name: "jane"}, 
{name: "imelda"}, 
{name: "john"},
{name: "jane"}
];

const data = Array.from(new Set(person.map(JSON.stringify))).map(JSON.parse);
console.log(data);

Answer 31

Here I found a simple solution for removing duplicates from an array of objects using reduce method. I am filtering elements based on the position key of an object

const med = [
  {name: 'name1', position: 'left'},
  {name: 'name2', position: 'right'},
  {name: 'name3', position: 'left'},
  {name: 'name4', position: 'right'},
  {name: 'name5', position: 'left'},
  {name: 'name6', position: 'left1'}
]

const arr = [];
med.reduce((acc, curr) => {
  if(acc.indexOf(curr.position) === -1) {
    acc.push(curr.position);
    arr.push(curr);
  }
  return acc;
}, [])

console.log(arr)

Answer 32

Continuing exploring ES6 ways of removing duplicates from array of objects: setting thisArg argument of Array.prototype.filter to new Set provides a decent alternative:

const things = [
  {place:"here",name:"stuff"},
  {place:"there",name:"morestuff"},
  {place:"there",name:"morestuff"}
];

const filtered = things.filter(function({place, name}) {

  const key =`${place}${name}`;

  return !this.has(key) && this.add(key);

}, new Set);

console.log(filtered);

However, it will not work with arrow functions () =>, as this is bound to their lexical scope.

Answer 33

es6 magic in one line... readable at that!

// returns the union of two arrays where duplicate objects with the same 'prop' are removed
const removeDuplicatesWith = (a, b, prop) => {
  a.filter(x => !b.find(y => x[prop] === y[prop]));
};

Answer 34

Source

JSFiddle

This will remove the duplicate object without passing any key.

uniqueArray = a => [...new Set(a.map(o => JSON.stringify(o)))].map(s => JSON.parse(s));

var objects = [{ 'x': 1, 'y': 2 }, { 'x': 2, 'y': 1 }, { 'x': 1, 'y': 2 }];

var unique = uniqueArray(objects);
console.log('Original Object',objects);
console.log('Unique',unique);

uniqueArray = a => [...new Set(a.map(o => JSON.stringify(o)))].map(s => JSON.parse(s));

    var objects = [{ 'x': 1, 'y': 2 }, { 'x': 2, 'y': 1 }, { 'x': 1, 'y': 2 }];

    var unique = uniqueArray(objects);
    console.log(objects);
    console.log(unique);

Answer 35

If you strictly want to remove duplicates based on one property, you can reduce the array into and object based on the place property, since the object can only have unique keys, you can then just get the values to get back to an array:

const unique = Object.values(things.thing.reduce((o, t) => ({ ...o, [t.place]: t }), {}))

Answer 36

I believe a combination of reduce with JSON.stringify to perfectly compare Objects and selectively adding those who are not already in the accumulator is an elegant way.

Keep in mind that JSON.stringify might become a performance issue in extreme cases where the array has many Objects and they are complex, BUT for majority of the time, this is the shortest way to go IMHO.

var collection= [{a:1},{a:2},{a:1},{a:3}]

var filtered = collection.reduce((filtered, item) => {
  if( !filtered.some(filteredItem => JSON.stringify(filteredItem) == JSON.stringify(item)) )
    filtered.push(item)
  return filtered
}, [])

console.log(filtered)

Another way of writing the same (but less efficient):

collection.reduce((filtered, item) => 
  filtered.some(filteredItem => 
    JSON.stringify(filteredItem ) == JSON.stringify(item)) 
      ? filtered
      : [...filtered, item]
, [])

Answer 37

 npm i lodash

 let non_duplicated_data = _.uniqBy(pendingDeposits, v => [v.stellarAccount, v.externalTransactionId].join());

Answer 38

The problem can be simplified to removing duplicates from the thing array.

You can implement a faster O(n) solution (assuming native key lookup is negligible) by using an object to both maintain unique criteria as keys and storing associated values.

Basically, the idea is to store all objects by their unique key, so that duplicates overwrite themselves:

const thing = [{ place: "here", name:"stuff" }, { place: "there", name:"morestuff" }, { place: "there", name:"morestuff" } ]

const uniques = {}
for (const t of thing) {
  const key = t.place + '$' + t.name  // Or whatever string criteria you want, which can be generified as Object.keys(t).join("$")
  uniques[key] = t                    // Last duplicate wins
}
const uniqueThing = Object.values(uniques)
console.log(uniqueThing)

Answer 39

const objectsMap = new Map();
const placesName = [
  { place: "here", name: "stuff" },
  { place: "there", name: "morestuff" },
  { place: "there", name: "morestuff" },
];
placesName.forEach((object) => {
  objectsMap.set(object.place, object);
});
console.log(objectsMap);

Answer 40

let data = [
  {
    'name': 'Amir',
    'surname': 'Rahnama'
  }, 
  {
    'name': 'Amir',
    'surname': 'Stevens'
  }
];
let non_duplicated_data = _.uniqBy(data, 'name');

Answer 41

If you don't mind your unique array being sorted afterwards, this would be an efficient solution:

things.thing
  .sort(((a, b) => a.place < b.place)
  .filter((current, index, array) =>
    index === 0 || current.place !== array[index - 1].place)

This way, you only have to compare the current element with the previous element in the array. Sorting once before filtering (O(n*log(n))) is cheaper than searching for a duplicate in the entire array for every array element (O(n²)).

Answer 42

str =[
{"item_id":1},
{"item_id":2},
{"item_id":2}
]

obj =[]
for (x in str){
    if(check(str[x].item_id)){
        obj.push(str[x])
    }   
}
function check(id){
    flag=0
    for (y in obj){
        if(obj[y].item_id === id){
            flag =1
        }
    }
    if(flag ==0) return true
    else return false

}
console.log(obj)

str is an array of objects. There exists objects having same value (here a small example, there are two objects having same item_id as 2). check(id) is a function that checks if any object having same item_id exists or not. if it exists return false otherwise return true. According to that result, push the object into a new array obj The output of the above code is [{"item_id":1},{"item_id":2}]

Answer 43

Have you heard of Lodash library? I recommend you this utility, when you don't really want to apply your logic to the code, and use already present code which is optimised and reliable.

Consider making an array like this

things.thing.push({place:"utopia",name:"unicorn"});
things.thing.push({place:"jade_palace",name:"po"});
things.thing.push({place:"jade_palace",name:"tigress"});
things.thing.push({place:"utopia",name:"flying_reindeer"});
things.thing.push({place:"panda_village",name:"po"});

Note that if you want to keep one attribute unique, you may very well do that by using lodash library. Here, you may use _.uniqBy

.uniqBy(array, [iteratee=.identity])

This method is like _.uniq (which returns a duplicate-free version of an array, in which only the first occurrence of each element is kept) except that it accepts iteratee which is invoked for each element in array to generate the criterion by which uniqueness is computed.

So, for example, if you want to return an array having unique attribute of 'place'

_.uniqBy(things.thing, 'place')

Similarly, if you want unique attribute as 'name'

_.uniqBy(things.thing, 'name')

Hope this helps.

Cheers!

Answer 44

• This solution is generic for any kind of object and checks for every (key, value) of the Object in the array.
• Using an temporary object as a hash table to see if the entire Object was ever present as a key.
• If the string representation of the Object is found then that item is removed from the array.

var arrOfDup = [{'id':123, 'name':'name', 'desc':'some desc'},
                {'id':125, 'name':'another name', 'desc':'another desc'},
                {'id':123, 'name':'name', 'desc':'some desc'},
                {'id':125, 'name':'another name', 'desc':'another desc'},
                {'id':125, 'name':'another name', 'desc':'another desc'}];

function removeDupes(dupeArray){
  let temp = {};
  let tempArray = JSON.parse(JSON.stringify(dupeArray));
  dupeArray.forEach((item, pos) => {
    if(temp[JSON.stringify(item)]){
      tempArray.pop();
    }else{
      temp[JSON.stringify(item)] = item;
    }
  });
 return tempArray;
}

arrOfDup = removeDupes(arrOfDup);

arrOfDup.forEach((item, pos) => {
  console.log(`item in array at position ${pos} is ${JSON.stringify(item)}`);
});

Answer 45

const uniqueElements = (arr, fn) => arr.reduce((acc, v) => {
    if (!acc.some(x => fn(v, x))) { acc.push(v); }
    return acc;
}, []);

const stuff = [
    {place:"here",name:"stuff"},
    {place:"there",name:"morestuff"},
    {place:"there",name:"morestuff"},
];

const unique = uniqueElements(stuff, (a,b) => a.place === b.place && a.name === b.name );

//console.log( unique );

[{
    "place": "here",
    "name": "stuff"
  },
  {
    "place": "there",
    "name": "morestuff"
}]

Answer 46

You can convert the array objects into strings so they can be compared, add the strings to a Set so the comparable duplicates will be automatically removed and then convert each of the strings back into objects.

It might not be as performant as other answers, but it's readable.

const things = {};

things.thing = [];
things.thing.push({place:"here",name:"stuff"});
things.thing.push({place:"there",name:"morestuff"});
things.thing.push({place:"there",name:"morestuff"});

const uniqueArray = (arr) => {

  const stringifiedArray = arr.map((item) => JSON.stringify(item));
  const set = new Set(stringifiedArray);

  return Array.from(set).map((item) => JSON.parse(item));
}

const uniqueThings = uniqueArray(things.thing);

console.log(uniqueThings);

Answer 47

Simple solution with ES6 'reduce' and 'find' array helper methods

Works efficiently and perfectly fine!

"use strict";

var things = new Object();
things.thing = new Array();
things.thing.push({
    place: "here",
    name: "stuff"
});
things.thing.push({
    place: "there",
    name: "morestuff"
});
things.thing.push({
    place: "there",
    name: "morestuff"
});

// the logic is here

function removeDup(something) {
    return something.thing.reduce(function (prev, ele) {
        var found = prev.find(function (fele) {
            return ele.place === fele.place && ele.name === fele.name;
        });
        if (!found) {
            prev.push(ele);
        }
        return prev;
    }, []);
}
console.log(removeDup(things));

Answer 48

For a readable and a simple solution searcher, her is my version:

    function removeDupplicationsFromArrayByProp(originalArray, prop) {
        let results = {};
        for(let i=0; i<originalArray.length;i++){
            results[originalArray[i][prop]] = originalArray[i];
        }
        return Object.values(results);
    }

Answer 49

My two cents here. If you know the properties are in the same order, you can stringify the elements and remove dupes from the array and parse the array again. Something like this:

var things = new Object();

things.thing = new Array();

things.thing.push({place:"here",name:"stuff"});
things.thing.push({place:"there",name:"morestuff"});
things.thing.push({place:"there",name:"morestuff"});
  
let stringified = things.thing.map(i=>JSON.stringify(i));
let unique =  stringified.filter((k, idx)=> stringified.indexOf(k) === idx)
                         .map(j=> JSON.parse(j))
console.log(unique);

Answer 50

• Removing Duplicates From Array Of Objects in react js (Working perfectly)

    let optionList = [];
        var dataArr = this.state.itemArray.map(item => {
            return [item.name, item]
        });
    var maparr = new Map(dataArr);
  
    var results = [...maparr.values()];
  
    if (results.length > 0) {
         results.map(data => {
         if (data.lead_owner !== null) {
              optionList.push({ label: data.name, value: 
              data.name });
         }
         return true;
       });
   }
   console.log(optionList)
  

Answer 51

You could use Set along with Filter method to accomplish this,

var arrObj = [{
  a: 1,
  b: 2
}, {
  a: 1,
  b: 1
}, {
  a: 1,
  b: 2
}];

var duplicateRemover = new Set();

var distinctArrObj = arrObj.filter((obj) => {
  if (duplicateRemover.has(JSON.stringify(obj))) return false;
  duplicateRemover.add(JSON.stringify(obj));
  return true;
});

console.log(distinctArrObj);

Set is a unique collection of primitive types, thus, won't work directly on objects, however JSON.stringify will convert it into a primitive type ie. String thus, we can filter.

If you want to remove duplicates based on only some particular key, for eg. key, you could replace JSON.stringify(obj) with obj.key

Answer 52

This is a single loop approach with a Set and some closures to prevent using declared variables outside function declarations and to get a short appearance.

const
    array = [{ place: "here", name: "stuff", n: 1 }, { place: "there", name: "morestuff", n: 2 }, { place: "there", name: "morestuff", n: 3 }],
    keys = ['place', 'name'],
    unique = array.filter(
        (s => o => (v => !s.has(v) && s.add(v))(keys.map(k => o[k]).join('|')))
        (new Set)
    );

console.log(unique);

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Answer 53

work for me

const uniqueArray = products.filter( (value,index) => {
  return index === products.findIndex( (obj) => { 
    return JSON.stringify(obj) === JSON.stringify(value);
  }) 
})

Answer 54

Here is another technique to find number of duplicate and and remove it easily from you data object. "dupsCount" is number of duplicate files count. sort your data first then remove. it will gives you fastest duplication remove.

  dataArray.sort(function (a, b) {
            var textA = a.name.toUpperCase();
            var textB = b.name.toUpperCase();
            return (textA < textB) ? -1 : (textA > textB) ? 1 : 0;
        });
        for (var i = 0; i < dataArray.length - 1; ) {
            if (dataArray[i].name == dataArray[i + 1].name) {
                dupsCount++;
                dataArray.splice(i, 1);
            } else {
                i++;
            }
        }

Answer 55

If you need an unique array based on multiple properties in the object you can do this with map and combining the properties of the object.

    var hash = array.map(function(element){
        var string = ''
        for (var key in element){
            string += element[key]
        }
        return string
    })
    array = array.filter(function(element, index){
        var string = ''
        for (var key in element){
            string += element[key]
        }
        return hash.indexOf(string) == index
    })

Answer 56

Generic for any array of objects:

/**
* Remove duplicated values without losing information
*/
const removeValues = (items, key) => {
  let tmp = {};

  items.forEach(item => {
    tmp[item[key]] = (!tmp[item[key]]) ? item : Object.assign(tmp[item[key]], item);
  });
  items = [];
  Object.keys(tmp).forEach(key => items.push(tmp[key]));

  return items;
}

Hope it could help to anyone.

Answer 57

Another way would be to use reduce function and have a new array to be the accumulator. If there is already a thing with the same name in the accumulator array then don't add it there.

let list = things.thing;
list = list.reduce((accumulator, thing) => {
    if (!accumulator.filter((duplicate) => thing.name === duplicate.name)[0]) {
        accumulator.push(thing);
    }
    return accumulator;
}, []);
thing.things = list;

I'm adding this answer, because I couldn't find nice, readable es6 solution (I use babel to handle arrow functions) that's compatible with Internet Explorer 11. The problem is IE11 doesn't have Map.values() or Set.values() without polyfill. For the same reason I used filter()[0] to get first element instead of find().

Answer 58

 var testArray= ['a','b','c','d','e','b','c','d'];

 function removeDuplicatesFromArray(arr){

 var obj={};
 var uniqueArr=[];
 for(var i=0;i<arr.length;i++){ 
    if(!obj.hasOwnProperty(arr[i])){
        obj[arr[i]] = arr[i];
        uniqueArr.push(arr[i]);
    }
 }

return uniqueArr;

}
var newArr = removeDuplicatesFromArray(testArray);
console.log(newArr);

Output:- [ 'a', 'b', 'c', 'd', 'e' ]

Answer 59

If you don't want to specify a list of properties:

function removeDuplicates(myArr) {
  var props = Object.keys(myArr[0])
  return myArr.filter((item, index, self) =>
    index === self.findIndex((t) => (
      props.every(prop => {
        return t[prop] === item[prop]
      })
    ))
  )
}

OBS! Not compatible with IE11.

Answer 60

here is my solution, it searches for duplicates based on object.prop and when it finds a duplicate object it replaces its value in array1 with array2 value

function mergeSecondArrayIntoFirstArrayByProperty(array1, array2) {
    for (var i = 0; i < array2.length; i++) {
        var found = false;
        for (var j = 0; j < array1.length; j++) {
            if (array2[i].prop === array1[j].prop) { // if item exist in array1
                array1[j] = array2[i]; // replace it in array1 with array2 value
                found = true;
            }
        }
        if (!found) // if item in array2 not found in array1, add it to array1
            array1.push(array2[i]);

    }
    return array1;
}

Answer 61

What about this:

function dedupe(arr, compFn){
    let res = [];
    if (!compFn) compFn = (a, b) => { return a === b };
    arr.map(a => {if(!res.find(b => compFn(a, b))) res.push(a)});
    return res;
}

Answer 62

If you find yourself needing to remove duplicate objects from arrays based on particular fields frequently, it might be worth creating a distinct(array, predicate) function that you can import from anywhere in your project. This would look like

const things = [{place:"here",name:"stuff"}, ...];
const distinctThings = distinct(things, thing => thing.place);

The distinct function can use any of the implementations given in the many good answers above. The easiest one uses findIndex:

const distinct = (items, predicate) => items.filter((uniqueItem, index) =>
    items.findIndex(item =>
        predicate(item) === predicate(uniqueItem)) === index);

Answer 63

You can use Object.values() combined with Array.prototype.reduce():

const things = new Object();

things.thing = new Array();

things.thing.push({place:"here",name:"stuff"});
things.thing.push({place:"there",name:"morestuff"});
things.thing.push({place:"there",name:"morestuff"});

const result = Object.values(things.thing.reduce((a, c) => (a[`${c.place}${c.name}`] = c, a), {})); 

console.log(result);

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Answer 64

Make Something simple. Fancy is good but unreadable code is useless. Enjoy :-)

var a = [
 {
  executiveId: 6873702,
  largePhotoCircle: null,
  name: "John A. Cuomo",
  photoURL: null,
  primaryCompany: "VSE CORP",
  primaryTitle: "Chief Executive Officer, President and Director"
 },
 {
  executiveId: 6873702,
  largePhotoCircle: null,
  name: "John A. Cuomo",
  photoURL: null,
  primaryCompany: "VSE CORP",
  primaryTitle: "Chief Executive Officer, President and Director"
 },
 {
  executiveId: 6873703,
  largePhotoCircle: null,
  name: "John A. Cuomo",
  photoURL: null,
  primaryCompany: "VSE CORP",
  primaryTitle: "Chief Executive Officer, President and Director",
 }
];

function filterDuplicate(myArr, prop) {
      // Format - (1)

      // return myArr.filter((obj, pos, arr) => {
      //     return arr.map(mapObj => mapObj[prop]).indexOf(obj[prop]) === pos;
      // });

      // Format - (2)
      var res = {};
      var resArr = [];
      for (var elem of myArr) {
        res[elem.executiveId] = elem;
      }
      for (let [index, elem] of Object.entries(res)) {
        resArr.push(elem);
      }
      return resArr;
  }
  
let finalRes = filterDuplicate(a,"executiveId");
console.log("finalResults : ",finalRes);

Answer 65

You can also create a generic function which will filter the array based on the object key you pass to the function

function getUnique(arr, comp) {

  return arr
   .map(e => e[comp])
   .map((e, i, final) => final.indexOf(e) === i && i)  // store the keys of the unique objects
   .filter(e => arr[e]).map(e => arr[e]); // eliminate the dead keys & store unique objects

 }

and you can call the function like this,

getUnique(things.thing,'name') // to filter on basis of name

getUnique(things.thing,'place') // to filter on basis of place

Answer 66

If you want to de-duplicate your array based on all arguments and not just one. You can use the uniqBy function of lodash that can take a function as a second argument.

You will have this one-liner:

 _.uniqBy(array, e => { return e.place && e.name })

Answer 67

function dupData() {
  var arr = [{ comment: ["a", "a", "bbb", "xyz", "bbb"] }];
  let newData = [];
  comment.forEach(function (val, index) {
    if (comment.indexOf(val, index + 1) > -1) {
      if (newData.indexOf(val) === -1) { newData.push(val) }
    }
  })
}

Answer 68

    function genFilterData(arr, key, key1) {
      let data = [];
      data = [...new Map(arr.map((x) => [x[key] || x[key1], x])).values()];
    
      const makeData = [];
      for (let i = 0; i < data.length; i += 1) {
        makeData.push({ [key]: data[i][key], [key1]: data[i][key1] });
      }
    
      return makeData;
    }
    const arr = [
    {make: "here1", makeText:'hj',k:9,l:99},
    {make: "here", makeText:'hj',k:9,l:9},
    {make: "here", makeText:'hj',k:9,l:9}]

      const finalData= genFilterData(data, 'Make', 'MakeText');
    
        console.log(finalData);

Answer 69

If you are using Lodash library you can use the below function as well. It should remove duplicate objects.

var objects = [{ 'x': 1, 'y': 2 }, { 'x': 2, 'y': 1 }, { 'x': 1, 'y': 2 }];
_.uniqWith(objects, _.isEqual);

Answer 70

We can leverage Javascript's Set object and Array's Filter function: For example:

// Example Array
const arr = [{ id: '1' }, { id: '2' }, { id: '1' }];
// Gather Unique Element Id's based on which you want to filter the elements.
const uniqIds = arr.reduce((ids, el) => ids.add(el.id), new Set());
// Filter out uniq elements.
const uniqElements = arr.filter((el) => uniqIds.delete(el.id));

console.log(uniqElements);

Answer 71

You can use for loop and condition to make it unique

const data = [
{ id: 1 },
{ id: 2 },
{ id: 3 },
{ id: 4 },
{ id: 5 },
{ id: 6 },
{ id: 6 },
{ id: 6 },
{ id: 7 },
{ id: 8 },
{ id: 8 },
{ id: 8 },
{ id: 8 }
];

const filtered= []

for(let i=0; i<data.length; i++ ){
    let isHasNotEqual = true
    for(let j=0; j<filtered.length; j++ ){
      if (filtered[j].id===data[i].id){
          isHasNotEqual=false
      }
    }
    if (isHasNotEqual){
        filtered.push(data[i])
    }
}
console.log(filtered);

/*
output
[ { id: 1 },
  { id: 2 },
  { id: 3 },
  { id: 4 },
  { id: 5 },
  { id: 6 },
  { id: 7 },
  { id: 8 } ]

*/

Answer 72

That's my solution by adding the actual array into key value object where the key is going to be the unique identify and the value could be any property of the object or the whole object.

Explanation: The main array with duplicate items will be transformed to a key/value object If the Id already exist in the unique object the value will be override. At the end just convert the unique object into an array.

getUniqueItems(array) {       
        const unique = {};
        // here we are assigning item.name but it could be a complete object.
        array.map(item => unique[item.Id] = item.name);
        // here you can transform your array item like {text: unique[key], value: key} but actually you can do what ever you want
        return Object.keys(unique).map(key => ({text: unique[key], value: key}));
      })
    );
  }

Answer 73

TypeScript function to filter an array to its unique elements where uniqueness is decided by the given predicate function:

function uniqueByPredicate<T>(arr: T[], predicate: (a: T, b: T) => boolean): T[] {
  return arr.filter((v1, i, a) => a.findIndex(v2 => predicate(v1, v2)) === i);
}

Without typings:

function uniqueByPredicate(arr, predicate) {
  return l.filter((v1, i, a) => a.findIndex(v2 => predicate(v1, v2)) === i);
}

Answer 74

This is simple way how to remove duplicity from array of objects.

I work with data a lot and this is useful for me.

const data = [{name: 'AAA'}, {name: 'AAA'}, {name: 'BBB'}, {name: 'AAA'}];
function removeDuplicity(datas){
    return datas.filter((item, index,arr)=>{
    const c = arr.map(item=> item.name);
    return  index === c.indexOf(item.name)
  })
}

console.log(removeDuplicity(data))

will print into console :

[[object Object] {
name: "AAA"
}, [object Object] {
name: "BBB"
}]

Answer 75

function filterDuplicateQueries(queries){
    let uniqueQueries = [];
     queries.forEach((l, i)=>{
        let alreadyExist = false;
        if(uniqueQueries.length>0){
            uniqueQueries.forEach((k, j)=>{
                if(k.query == l.query){
                    alreadyExist = true;
                }
            });
        }
        if(!alreadyExist){
           uniqueQueries.push(l)
        }
    });

Answer 76

var things = new Object();

things.thing = new Array();

things.thing.push({place:"here",name:"stuff"});
things.thing.push({place:"there",name:"morestuff"});
things.thing.push({place:"there",name:"morestuff"});
console.log(things);
function removeDuplicate(result, id) {
    let duplicate = {};
    return result.filter(ele => !duplicate[ele[id]] &&                   (duplicate[ele[id]] = true));
}
let resolverarray = removeDuplicate(things.thing,'place')
console.log(resolverarray);

Answer 77

Here is a solution using new filter function of JavaScript that is quite easy . Let's say you have an array like this.

var duplicatesArray = ['AKASH','AKASH','NAVIN','HARISH','NAVIN','HARISH','AKASH','MANJULIKA','AKASH','TAPASWENI','MANJULIKA','HARISH','TAPASWENI','AKASH','MANISH','HARISH','TAPASWENI','MANJULIKA','MANISH'];

The filter function will allow you to create a new array, using a callback function once for each element in the array. So you could set up the unique array like this.

var uniqueArray = duplicatesArray.filter(function(elem, pos) {return duplicatesArray.indexOf(elem) == pos;});

In this scenario your unique array will run through all of the values in the duplicate array. The elem variable represents the value of the element in the array (mike,james,james,alex), the position is it's 0-indexed position in the array (0,1,2,3...), and the duplicatesArray.indexOf(elem) value is just the index of the first occurrence of that element in the original array. So, because the element 'james' is duplicated, when we loop through all of the elements in the duplicatesArray and push them to the uniqueArray, the first time we hit james, our "pos" value is 1, and our indexOf(elem) is 1 as well, so James gets pushed to the uniqueArray. The second time we hit James, our "pos" value is 2, and our indexOf(elem) is still 1 (because it only finds the first instance of an array element), so the duplicate is not pushed. Therefore, our uniqueArray contains only unique values.

Here is the Demo of above function.Click Here for the above function example